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Title says it all. I'm guessing it shouldn't be possible for an estimator to be UMVUE and yet not consistent. Is there a counter-example to this? Or is there a proof that it can't happen?

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One counterexample is when there's no consistent estimator. Suppose $X_i\sim N(\mu_i,1)$ where $\mu_i$ are all distinct unknown parameters. The UMVUE of any $\mu_i$ is $X_i$, but it's not consistent.

You can make matters worse. Suppose you a have single $X\sim N(\theta,1)$ together with $Y_i\sim\text{Bernoulli}(p)$ with $\mathrm{logit}\,p=\theta$ for $i=1,\dots,n$. Then $\mathrm{logit}\, \bar Y_n$ is a consistent estimator for $\theta$ as $n\to\infty$, but no function of $Y_i$ is unbiased for $\theta$. This means the UMVUE of $\theta$ is just $X$, which is not consistent.

However, for iid $X_i$, $i=1,\dots,n$, if we assume there exists some unbiased estimator $\tilde\theta$ that has finite variance for all $n$ greater than some $n_0$, the MVUE must be consistent. We can divide the $n$ observations into $[n/n_0]$ blocks of size at least $n_0$ and take the average of $\tilde\theta$ for each block. If $\sigma^2$ is the variance of each block-specific $\tilde\theta$ then the variance of the average is $\sigma^2/[n/n_0]$, which goes to zero as $n\to\infty$. So there is a consistent unbiased estimator and so the MVUE is also consistent.

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